Aniline reactions and preparation

Preparation. Nitrobenzene is reduced completely — six "hydrogen equivalents" replace the three oxygens of $-NO_2$ with two hydrogens on nitrogen:

where: $[H]$ is the reducing hydrogen supplied in situ by a metal–acid couple ($Sn+HCl$, $Fe+HCl$, or catalytically $H_2/Ni$). With $Sn/HCl$ the immediate product is anilinium chloride, $C_6H_5NH_3^+Cl^-$; a subsequent $NaOH$ wash liberates the free amine. This two-stage picture — reduce, then basify — is worth stating explicitly in any preparation question.

Physical properties. Aniline is a colourless, oily liquid that darkens to brown on standing in air and light (slow aerial oxidation), boils at $184\,^\circ C$, and is only slightly water-soluble: the $-NH_2$ group can hydrogen-bond to water, but the large non-polar phenyl ring dominates and limits mixing. It dissolves freely in ethanol and ether.

One lone pair, two jobs. The nitrogen lone pair is the key to almost every reaction below. Free to donate into an acid's proton, it makes aniline a (weak) base; delocalised into the aromatic ring by resonance, it makes the ring strongly activated toward electrophiles. Any substituent placed on nitrogen (acetyl, alkyl, benzoyl) partially withdraws that lone pair from one job or the other, and that trade-off explains the reaction pattern below.

Aniline's reactivity is a story of one lone pair being pulled in different directions, not a formula to derive — so the logic, not an equation, carries the marks.

Basicity is weak because the lone pair is shared with the ring. In ammonia or an aliphatic amine the nitrogen lone pair sits entirely on nitrogen and is fully available to grab a proton. In aniline, resonance delocalises part of that lone pair into the benzene ring (donating electron density to the ortho and para positions), so less electron density remains localised on nitrogen. The result is a weaker base than ammonia, even though aniline is still basic enough to form salts with strong acids like $HCl$ and $H_2SO_4$.

Acylation and benzoylation trade basicity for ring protection. When acetic anhydride or benzoyl chloride reacts with the $-NH_2$ nitrogen, an acyl group is bonded to nitrogen. The nitrogen lone pair is now shared with the new carbonyl too, so the amide product (acetanilide, benzanilide) is a much weaker base and, crucially, cannot be protonated by dilute acid — which matters directly for nitration, worked through below.

Diazotisation needs cold conditions because the product is only moderately stable. At $0$–$5\,^\circ C$, nitrous acid (generated in situ from $NaNO_2 + HCl$) converts the free $-NH_2$ into a diazonium ion, $-N_2^+$. Aryl diazonium salts are more stable than aliphatic ones (the positive charge is stabilised by the adjacent ring), but they still decompose above about $5\,^\circ C$, releasing $N_2$ and giving phenol — so the low-temperature condition is not a minor detail but the reason the reaction succeeds at all.

The carbylamine reaction is diagnostic because it needs a free, primary $-NH_2$. Only a primary amine (one $-NH_2$ directly on the parent group, with no other substituent) reacts with $CHCl_3$/alcoholic $KOH$ to give the foul-smelling isocyanide. Secondary or acylated amines give no such smell, which is exactly why this reaction is used as a specific test.

Problem: Nitrobenzene is first reduced to aniline. Some of the aniline is diazotised directly; the rest is acetylated and then nitrated. Write the equations for each stage, and explain why the two routes give such different ring products.

Step 1 (reduction). $C_6H_5NO_2 + 3Sn + 6HCl \rightarrow C_6H_5NH_3^+Cl^- + 3SnCl_2 + 2H_2O$, followed by $NaOH$ to liberate free aniline, $C_6H_5NH_2$.

Step 2 (direct diazotisation). The free aniline is dissolved in excess dilute $HCl$, cooled to $0$–$5\,^\circ C$, and treated with $NaNO_2$ solution: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5\,^\circ C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$. This gives benzenediazonium chloride — the free $-NH_2$'s lone pair is what makes this possible, and the low temperature keeps the diazonium ion from decomposing before use.

Step 3 (acetylation then nitration). Separately, aniline is acetylated: $C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$. Under the strongly acidic nitrating mixture, free aniline would be protonated to the meta-directing anilinium ion, giving unwanted meta-nitroaniline; the acetamido group cannot be protonated, so it stays a moderate, o,p-directing activator throughout. Nitration of acetanilide therefore gives predominantly p-nitroacetanilide, and hydrolysis afterward regenerates the amine as p-nitroaniline.

Why the routes differ: the diazotisation route needs the lone pair fully free and available on nitrogen (so no protection is used), while the nitration route needs that same lone pair temporarily dampened and un-protonatable (so protection by acetylation is essential). Same starting material, opposite treatment of the lone pair, because the two target reactions make opposite demands on it.

• Assuming aniline is more basic than ammonia because "amines are basic" — aniline is markedly weaker due to lone-pair delocalisation into the ring; do not generalise from aliphatic amines.
• Skipping the $0$–$5\,^\circ C$ condition in diazotisation — above this range the diazonium salt decomposes to phenol before it can be isolated or used further.
• Treating acetylation as an end in itself rather than as protection: its real exam purpose is temporarily reducing the ring-activating, meta-avoiding power of $-NH_2$ so nitration is controllable.
• Mixing up acetylation (acid anhydride/chloride, gives an amide, retains the C–N bond with an acyl group added) with alkylation (alkyl halide, gives a substituted amine, no carbonyl introduced).

Ask the tutor: want to walk through why acetanilide is still o,p-directing but much less strongly activating than free aniline, using resonance structures?